September 22, 2019, 04:41:06 AM
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Topic: How do we predict how SbF5 shifts the autodissociation equilibrium of liquid HF?  (Read 3656 times)

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Offline cncbmb

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How do we predict how SbF5 shifts the autodissociation equilibrium of liquid HF?

Context:

Page 16, 2008 Chemistry Olympiad Prep Problems:
http://www.icho.hu/Files/prep_problems_icho40_0521.pdf

h) Show how SbF5 shifts the autodissociation equilibrium of liquid HF.
The shift in the autodissociation also implies an important change in the Brønsted acidity of the solvent. In fact, the degree of solvation of the proton produced from the autodissociation essentially affects the Brønsted acidity of the solvent.

In water HF behaves as a medium-strength acid and dissociates only partially. The most important reactions determining the equilibrium properties of the solution are the following:
HF + H2O 􀁕 H3O+ + F– (1)
HF + F– 􀁕 HF2– (2)
The equilibrium constants of the two equilibria are
K1 = 1.1·10^–3
K2 = 2.6·10^–1

How do we predict how SbF5 shifts the autodissociation equilibrium of liquid HF?

Apparently, the answer is 2HF+SbF5->SbF6- + H2F+
on page 60 of that above pdf file but I don't know why this is better than other reactions like

HF+SbF5->SbF6+ + F-? Why is 2HF+SbF5- >SbF6- + H2F+ the best equation for this?

Also, how can we hide text and type math in prettier format than html text?

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